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July 10th, 2014, 09:40 AM   #1
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relationship between derivative of function and derivative of inverse

from dx/dy = 1/y', derive d^2/dy^2 = -y''/(y')^3 and
d^3x/dy^3 = 3(y'')^2-y'y''/(y')^5.
I have no idea where to start please explain. Also can you please explain why the second and higher derivatives are d^nx/dy^n and not dx^n/dy^n?
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July 10th, 2014, 03:38 PM   #2
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$\newcommand{\d}{\mathbb{d}}$
$$\text{From} \qquad \frac{\d x}{\d y} = \frac{1}{y^\prime} \qquad \text{derive} \qquad \frac{\d^2x}{\d y^2} \qquad \text{and} \qquad \frac{\d^3x}{\d y^3}$$
We just need to differentiate, but note that $y$ is a function of $x$, and in particular $y^\prime$ is a function of $x$. Thus to differentiate with respect to $y$ we need to use the chain rule by writing
\begin{align*}
\frac{\d x}{\d y} &= \frac{1}{y^\prime(x(y))} &&= {y^\prime}^{-1}(x(y)) \\
\frac{\d^2x}{\d y^2} &= -{y^\prime}^{-2}(x(y)) y^{\prime\prime}(x(y)) \frac{\d x}{\d y} &&= -{y^\prime}^{-3}(x(y)) y^{\prime\prime}(x(y)) &&= -\frac{y^{\prime\prime}}{{y^\prime}^{3}} \\
\frac{\d^3x}{\d y^3} &= 3{ y^{ \prime\prime } (x(y)) }{ y^\prime(x(y)) }^{-4}{ y^{ \prime\prime }(x(y)) }\frac{\d x}{\d y} + -{ y^\prime(x(y)) }^{-3}y^{ \prime\prime\prime }(x(y))\frac{\d x}{\d y}
&&= 3{y^{ \prime\prime }(x(y))}^2{ y^\prime(x(y)) }^{-5} + -{ y^\prime(x(y)) }^{-4}y^{ \prime\prime\prime }(x(y))
&&= \frac{3{y^{ \prime\prime }}^2 + y^\prime y^{ \prime\prime\prime }}{ { y^\prime }^{5} }

\end{align*}

We write $\frac{\d^2 y}{\d x^2}$ because $\frac{\d}{\d x}$ is an operator. We are not squaring $\frac{\d y}{\d x}$, but doing the operation of differentiation twice on $y$.
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July 10th, 2014, 06:57 PM   #3
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Thank you very much!
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